CN104199030B - rotary synthetic aperture radar frequency domain imaging method - Google Patents

Abstract

The present invention relates to a kind of rotary synthetic aperture radar frequency domain imaging method, the present invention proposes ROSAR frequency domain imaging method.The method utilizes principle in phase point directly ROSAR echo-signal to be transformed to two-dimensional frequency, therefore can retain echo character well, by constructing corresponding frequency domain decoupling function and frequency matching function, finally realizes range curvature correction and target scene reconstruct.Simulation result shows, the method can quickly bend by correction distance, and can effectively overcome orientation to mismatch problems.

Description

Rotary synthetic aperture radar frequency domain imaging method

Technical field

The present invention relates to a kind of rotary synthetic aperture radar frequency domain imaging method, belong to synthetic aperture radar neck Territory.

Background technology

Synthetic aperture radar (SAR) has round-the-clock, round-the-clock, remote ability to work, extensively uses In fields such as landforms imaging, moving object detection, hot spot region supervision.In traditional SAR imaging pattern, Mainly include strip-type imaging, bunching type imaging, scan-type imaging and circumferentially imaging.Rotary synthesis hole Footpath radar (ROSAR), as a kind of new imaging pattern, is the important supplement part of satellite-borne SAR and carried SAR, It is increasingly becoming a new study hotspot in the nearest more than ten years.This pattern has that areas imaging is wide, revisiting period is short, It is applicable to the advantages such as short distance imaging, both can apply to the low hollow panel such as unmanned plane, helicopter, it is also possible to should For the ground surface platform such as lorry, circuit orbit.ROSAR is also referred to as Circular test SAR, circular arc SAR, omnidirectional SAR With circumference SAR.

Existing ROSAR formation method is based primarily upon oblique distance Taylor series expansion, ignores four times and high-order term, so The methods such as rear combination RD, ECS, SPECAN carry out imaging to echo data.Obtained by Taylor series expansion Oblique distance be approximately parabola, corresponding doppler values is the linear function of slow time, and doppler frequency rate is normal Numerical value.And the echo Doppler of actual ROSAR is nonlinear change, doppler frequency rate is not constant value, because of This utilizes the frequency modulation rate obtained by Taylor series approximation to carry out pulse pressure, it will cause orientation to mismatch.Time existing Territory convolution formation method does not do Taylor series expansion, directly carries out time domain coupling imaging, it is possible to the preferably side of completing Position is to coupling.But in distance in the case of high-resolution, range curvature can not be ignored, corresponding time domain bends Bearing calibration will be significantly greatly increased amount of calculation.

Summary of the invention

For the problems referred to above, the present invention utilizes principle in phase point that echo-signal directly carries out frequency domain transform, it is thus achieved that Bidimensional frequency-domain expression, and then structure frequency domain decoupling function, distance are to adaptation function and azimuth match function, Then on frequency domain, realize range curvature correction and target scene reconstruct.Carrying out range curvature correction when, The present invention also analyzes the range migration impact of ROSAR.The simulation results show effectiveness of method.

Accompanying drawing explanation

The exemplary embodiment of the present invention it is more fully described, the above and other side of the present invention by referring to accompanying drawing Face and advantage will become the clearest, in the accompanying drawings:

Fig. 1 is the ROSAR geometric model schematic diagram of the rotary synthetic aperture radar frequency domain imaging method of the present invention；

Fig. 2 (a) is t_r-t_aTerritory echo data schematic diagram；Fig. 2 (b) is t_r-f_aTerritory echo data schematic diagram；

Fig. 3 is ROSAR top view；

Fig. 4 is Δ R (f_amax；R_G) change curve；

Fig. 5 is ROSAR Irnaging procedures figure；

Fig. 6 (a) is schematic diagram before range curvature correction；Fig. 6 (b) is schematic diagram after range curvature correction；

Fig. 7 (a) is traditional pulse pressure result schematic diagram；Fig. 7 (b) is the pulse pressure result schematic diagram of the present invention；

Fig. 8 is that orientation is to profile.

Detailed description of the invention

Hereinafter, it is more fully described the present invention, various enforcements shown in the drawings now with reference to accompanying drawing Example.But, the present invention can implement in many different forms, and should not be construed as limited to explain at this The embodiment stated.On the contrary, it is provided that these embodiments make the disclosure will be thoroughly and completely, and by the present invention Scope be fully conveyed to those skilled in the art.

Hereinafter, the exemplary embodiment of the present invention it is more fully described with reference to the accompanying drawings.

Fig. 1 is the geometric model of ROSAR, and radar antenna is fixed on helicopter wing, or is arranged on unmanned On machine or on the rigid support of lorry.The present invention is as a example by ROSAR typical platform helicopter, it is assumed that carrier aircraft The height of platform is H, and antenna acts as uniform circular motion with angular velocity omega along with wing one, the most periodically Launching signal, backscattering echo returns to antenna through a fixed response time, then enters the echo-signal received Row storage.Antenna forms an annular and irradiates scene, point target P in scene on ground_nTo center of rotation axle Ground distance (abbreviation distance) be R_G, a length of L of rotor, orientation is γ to beam angle, and the angle of pitch is β, pitching is ε to beam angle.X-axis direction faces out with being defined as parallel to, and z-axis direction is defined as vertically Upwards, y-axis is perpendicular to x-z-plane on ground.Owing to antenna circles, therefore the present invention uses cylinder to sit Mark system (r,z).Not losing a characteristic of stock, when antenna phase central rotation to A point, wing is parallel with x-axis, Order anglec of rotation size now is 0 degree, and when antenna rotates to other arbitrfary point B, anglec of rotation size is ω t_a。 As seen from the figure, the position of B point is (L, ω t_a, H), it is assumed that point target P_nPosition be (R_G, ω t_n, 0), Then antenna to the distance expression formula of point target is

$R(t_a;R_G)=\sqrt{L^2-L{R}_G\cos \left[\mathrm{\ω}(t_a-t_n)\right]+R_G^2+H^2}---\left(1\right)$

The signal of radar emission is linear FM signal, then backscattering echo signal through fundamental frequency convert, away from The slow time domain in m-orientation (t when fast_r-t_aTerritory) it is represented by

$\begin{array}{c}{s}_n(t_r,t_a;R_G)=a_r[t_r-\frac{2R(t_a;R_G)}{c}]a_a\left(t_a\right)\exp \left\{\mathrm{j\π}{K}_r{[t_r-\frac{2R(t_a;R_G)}{c}]}^2\right\}\\ \×\exp [-j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(t_a;R_G)]\end{array}---\left(2\right)$

Wherein, a_r() represent that distance is to window function, a_a() represent that orientation is to window function, K_rFor signal frequency modulation rate, λ For launching the wavelength of signal, c represents the light velocity.

1. frequency domain imaging method

To s_n(t_r, t_a；R_G) carry out orientation to frequency domain (f_aTerritory) conversion, can obtain

$s_n(t_r,f_a;R_G)={\∫}_{t_n-{\mathrm{\α}}_S/2\mathrm{\ω}}^{t_n+{\mathrm{\α}}_S/2\mathrm{\ω}}{s}_n(t_r,t_a;R_G)\exp (-j2\mathrm{\π}{f}_a{t}_a){\mathrm{dt}}_a---\left(3\right)$

Wherein, t_nIt it is point target P_nThe position time point upwards in orientation, α_SIt is that orientation is to corresponding to synthetic aperture S Synthetic aperture angle.α_SExpression formula be

${\mathrm{\α}}_S=\mathrm{\γ}(1-\frac{L}{R_G})---\left(4\right)$

Integrand form in formula (3) is extremely complex, and this makes the formula that is difficult to obtain (3) integral result, to this end, The present invention uses principle in phase point to ask for the integral result of formula (3).The precondition of utilization principle in phase point is The amplitude of integrand is constant or tempolabile function, and phase place changes much faster, and rate of change is to change. Amplitude in formula (3) is gradual, and phase place is expressed as

Phase place in formula (5) is fast-changing in integrating range, and rate of change is to change, and therefore can utilize Principle in phase point solves formula (3).

By formula (5) to slow time t_aDerivation, and function is 0 after making derivation, can be in the hope of

$\sin \left[\mathrm{\ω}(t_a^{*}-t_n)\right]=-\frac{\mathrm{\λ}}{2L{R}_G\mathrm{\ω}}{f}_a\sqrt{L^2+R_G^2+H^2-[\frac{{\mathrm{\λ}}^2}{2{\mathrm{\ω}}^2}{f}_a^2+\sqrt{\frac{{\mathrm{\λ}}^4}{4{\mathrm{\ω}}^4}{f}_a^4-\frac{{\mathrm{\λ}}^2}{{\mathrm{\ω}}^2}(L^2+R_G^2+H^2)f_a^2+4L^2{R}_G^2}]}---\left(6\right)$

Wherein,It is in phase point.As ω (t_a-t_nDuring)=pi/2, oblique distance isAnd makeThen (6) can be simplified, and then obtain

$\begin{array}{c}{t}_a^{*}=t_n-\frac{1}{\mathrm{\ω}}\arcsin \left[\frac{1}{2L{R}_G}\stackrel{\‾}{f_a}\sqrt{R_{\max }^2-(\frac{1}{2}{\stackrel{\‾}{f}}_a^2+\sqrt{\frac{1}{4}{\stackrel{\‾}{f}}_a^4-R_{\max }^2{\stackrel{\‾}{f}}_a^2+{4L}^2{R}_G^2})}\right]\\ =t_n-{\stackrel{\‾}{t_a}}^{*}\end{array}---\left(7\right)$

According to principle in phase point, formula (7) is substituted into formula (3), can obtain

$\begin{array}{c}{s}_n(t_r,t_a;R_G)=a_r[t_r-\frac{2R(f_a;R_G)}{c}]a_a\left(t_a^{*}\right)\exp \left\{\mathrm{j\π}{K}_e(f_a;R_G){[t_r-\frac{2R(f_a;R_G)}{c}]}^2\right\}\\ \×\exp [-j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(f_a;R_G)]\exp \left(j2\mathrm{\π}{f}_a{\stackrel{\‾}{t_a}}^{*}\right)\exp (-j2\mathrm{\π}{f}_a{t}_n)\end{array}---\left(8\right)$

Wherein, R (f_a；R_G) it is that oblique distance R is at f_aExpression formula on territory.Work as t_aTerritory transforms to f_aDuring territory, launch signal Frequency modulation rate K_rAlso can change, make K_rAt f_aK it is expressed as on territory_e(f_a；R_G).Be given individually below R(f_a；R_G) and K_e(f_a；R_G) expression.

Formula (7) is substituted into formula (1), tries to achieve R (f_a；R_G) expression formula be

$R(f_a;R_G)=\sqrt{R_{\max }^2-(\frac{1}{2}{\stackrel{\‾}{f}}_a^2+\sqrt{\frac{1}{4}{\stackrel{\‾}{f}}_a^4-R_{\max }^2{\stackrel{\‾}{f}}_a^2+{4L}^2{R}_G^2})}---\left(9\right)$

Work as f_aWhen=0, R (f can be obtained_a；R_G) constant term beThis constant term It it is point target P_nFormula (9) therefore can be rewritten as by the beeline in oblique distance face

R(f_a；R_G)=R_C+ΔR(f_a；R_G) (10)

Wherein, Δ R (f_a；R_G) it is R (f_a；R_G) once and high-order term.

K can be obtained by analysis below_e(f_a；R_G).As shown in Fig. 2 (a), at t_r-t_aA single point target in territory Echo data along fast time t_rVertical distribution is on certain string storage position.Work as t_aTerritory transforms to f_aDuring territory, return The storage position of wave datum becomes skew lines, shown in solid as in Fig. 2 (b).It is f for tranmitting frequency_cLetter Number, antenna rotates to a certain angle, θ_t=ω t_a, doppler values now is

$F_a=-\frac{2L{R}_G\mathrm{\ω}\sin \left({\mathrm{\θ}}_t\right)f_c}{\mathrm{cR}\left({\mathrm{\θ}}_t\right)}---\left(11\right)$

Wherein, R (θ_t) it is the R (t in formula (1)_a；R_G).It is linear FM signal owing to launching signal, when launching frequency Rate becomes f=f_c+ Δ f (Δ f=K_rΔt_r), anglec of rotation θ_tCorresponding doppler values is

$F_a^{\′}=-\frac{2L{R}_G\mathrm{\ω}\sin \left({\mathrm{\θ}}_t\right)(f_c+\mathrm{\Δf})}{\mathrm{cR}\left({\mathrm{\θ}}_t\right)}---\left(12\right)$

Now, due to the change of tranmitting frequency, storage position is transferred to a little 2 by putting 1, i.e. for a single point target Saying, the storage position of echo data is at t_r-f_aIt territory is skew lines.

It is f for tranmitting frequency_c+ Δ f signal, doppler values F_aCorresponding storage data are another echoes Received data, i.e. echo data received for aerial position B ' in Fig. 3, now the anglec of rotation is θ_t-Δθ_t.Corresponding doppler values expression formula is

$F_a=-\frac{2L{R}_G\mathrm{\ω}\sin ({\mathrm{\θ}}_t-\mathrm{\Δ}{\mathrm{\θ}}_t)(f_c+\mathrm{\Δf})}{\mathrm{cR}({\mathrm{\θ}}_t-\mathrm{\Δ}{\mathrm{\θ}}_t)}---\left(13\right)$

Understand, due to t according to above analysis_aTerritory transforms to f_aTerritory, the transformation required for same frequency difference DELTA f Time is by different.(the T in such as Fig. 2 (a) on same slow time point_a), tranmitting frequency is by f_cConversion To f_c+ Δ f required time is Δ t_r(Δt_r=Δ f/K_r), but for same doppler values F_a, tranmitting frequency By f_cTransform to f_c+ Δ f required time Δ t '_r(Δt′_r=Δ f/K_e(f_a；R_G)) it is

$\mathrm{\Δ}{t}_r^{\′}={\mathrm{\Δt}}_r-\frac{2\mathrm{\ΔR}\left({\mathrm{\θ}}_t\right)}{c}---\left(14\right)$

Wherein, Δ R (θ_t)=R (θ_t)-R(θ_t-Δθ_t).In order to eliminate intermediate variable Δ f, need Δ R (θ_t) be expressed as The expression formula of Δ f, can obtain according to formula (1)

$\mathrm{\ΔR}\left({\mathrm{\θ}}_t\right)=R\left({\mathrm{\θ}}_t\right)-R({\mathrm{\θ}}_t-\mathrm{\Δ}{\mathrm{\θ}}_t)=\frac{{\mathrm{LR}}_G}{R\left({\mathrm{\θ}}_t\right)}\sin \left({\mathrm{\θ}}_t\right)\sin \left(\mathrm{\Δ}{\mathrm{\θ}}_t\right)---\left(15\right)$

Can obtain according to formula (11) and formula (13) again

$\sin \left(\mathrm{\Δ}{\mathrm{\θ}}_t\right)=\tan \left({\mathrm{\θ}}_t\right)[\frac{\mathrm{\ΔR}\left({\mathrm{\θ}}_t\right)}{R\left({\mathrm{\θ}}_t\right)}+\frac{\mathrm{\Δf}}{f_c}]---\left(16\right)$

Formula (16) is substituted into formula (15), it is thus achieved that Δ R (θ_t) about the expression formula of Δ f be

$\mathrm{\ΔR}\left({\mathrm{\θ}}_t\right)=\frac{{\mathrm{LR}}_G\sin \left({\mathrm{\θ}}_t\right)\tan \left({\mathrm{\θ}}_t\right)}{R^2\left({\mathrm{\θ}}_t\right)-{\mathrm{LR}}_G\sin \left({\mathrm{\θ}}_t\right)\tan \left({\mathrm{\θ}}_t\right)}\frac{\mathrm{\Δf}}{f_c}R\left({\mathrm{\θ}}_t\right)---\left(17\right)$

Variable in formula (17) is θ_t(θ_t=ω t_a), t in formula to be utilized (7)_aWith f_aRelation, by formula (17) Transform to f_aTerritory, corresponding θ_tIt is transformed into θ_f.Finally utilize formula (14) and formula (17), after eliminating intermediate variable Δ f Can obtain

Due to Δ t_r=Δ f/K_r, Δ t '_r=Δ f/K_e(f_a；R_G), can obtain according to formula (16) and formula (17)

$\frac{1}{K_e(f_a;R_G)}=\frac{1}{K_r}-\frac{{2\mathrm{LR}}_G\sin \left({\mathrm{\θ}}_f\right)\tan \left({\mathrm{\θ}}_f\right)}{R^2\left({\mathrm{\θ}}_f\right)-{\mathrm{LR}}_G\sin \left({\mathrm{\θ}}_f\right)\tan \left({\mathrm{\θ}}_f\right)}\frac{\mathrm{\λ}}{c^2}R\left({\mathrm{\θ}}_f\right)---\left(18\right)$

So, oblique distance and frequency modulation rate are just obtained at f_aExpression formula on territory, by the R (f of formula (10)_a；R_G) and formula (18) K_e(f_a；R_G) substitute into formula (8), i.e. can get echo-signal at t_r-f_aExpression formula on territory s_n(t_r, f_a；R_G)。

After orientation is to frequency domain transform, recycle principle in phase point, to s_n(t_r, f_a；R_G) carry out distance to Frequency domain (f_rTerritory) conversion, can be in the hope of

$\begin{array}{c}{s}_n(f_r,f_a;R_G)=a_r\left[\frac{f_r}{K_e(f_a;R_G)}\right]a_a\left(t_a^{*}\right)\exp [-\mathrm{j\π}\frac{f_r^2}{K_e(f_a;R_G)}]\\ \×\exp [-j\frac{4\mathrm{\π}}{c}\mathrm{\ΔR}(f_a;R_G)f_r]\exp (-j\frac{4\mathrm{\π}}{c}{R}_C{f}_r)\\ \×\exp [-j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(f_a;R_G)]\exp \left(j2\mathrm{\π}{f}_a{\stackrel{\‾}{t_a}}^{*}\right)\exp (-j2\mathrm{\π}{f}_a{t}_n)\end{array}---\left(19\right)$

Above formula is the two-dimensional frequency expression formula of ROSAR echo-signal.By directly do orientation to distance to Frequency domain transform, preferably remain the spectrum signature of ROSAR echo-signal, it is possible to effectively solve oblique distance R Produced by Taylor series expansion, orientation is to mismatch problems.

Range migration will affect the imaging effect of ROSAR, and the distance under ROSAR pattern to be analyzed is moved Dynamic impact, specifically includes range walk, range curvature and the impact of distance space-variant.In the case of one, ROSAR The irradiation beam direction of antenna is mutually perpendicular to the movement velocity direction of antenna, is similar to the positive side-looking of stripmap SAR Pattern, the present invention is also adopted by this pattern, therefore need not consider the impact of range walk.In formula (19) second Individual exponential term is the azimuthrange coupling item caused by range curvature, and therefore the decoupling function of correction distance bending is

$H_1(f_r,f_a;R_G)=\exp \left[j\frac{4\mathrm{\π}}{c}\mathrm{\ΔR}(f_a;R_G)f_r\right]---\left(20\right)$

Wherein, Δ R (f_a；R_G) degree of crook with R_GChange and change, i.e. oblique distance R is at f_aSpace-variant is there is in territory Property.

The synthetic aperture angle provided according to formula (4), can be in the hope of a certain distance R_GCorresponding is the most general Le value is

$f_{a\max }=-\frac{2L{R}_G\mathrm{\ω}\sin \left(\frac{{\mathrm{\α}}_S}{2}\right)}{\mathrm{\λ}\sqrt{L^2-2L{R}_G\cos \left(\frac{{\mathrm{\α}}_S}{2}\right)+R_G^2+H^2}}---\left(21\right)$

Can be in the hope of oblique distance R at f according to formula (10) and formula (21)_aThe ultimate value of maximum deflection value in territory

$\underset{R_G\→\∞}{\mathrm{lin}}\mathrm{\ΔR}(f_{a\max };R_G)=L[1-\cos \left(\frac{\mathrm{\γ}}{2}\right)]---\left(22\right)$

Fig. 4 is differently away from R_GOblique distance maximum deflection value change curve, with R_GIncrease, this curve tends to One fixed value.As seen from the figure, R is worked as_GAfter certain value (such as 500m), oblique distance maximum deflection value Change tends towards stability, and corresponding scene is remote, proximal edge degree of crook reaches unanimity, in the case of one, and ROSAR The distance value irradiating scene center is relatively big (such as 1000m, 2000m etc.), now can ignore space-variant Impact.In the case of not considering space-variant, one distance choosing scene center is reference distance, then Range curvatures all in scene are unified correction.

After range curvature correction, in order to carry out distance to pulse pressure, distance adaptation function need to be constructed as follows

$H_2(f_r,f_a;R_G)=\exp \left[\mathrm{j\π}\frac{f_r^2}{K_e(f_a;R_G)}\right]---\left(23\right)$

Original fundamental frequency echo data is carried out bidimensional frequency domain transform, data are transformed into f_r-f_aTerritory, then utilizes Formula (20) and formula (23) realize range curvature correction and distance to pulse pressure.In order to complete orientation to pulse pressure, structure side Position is as follows to adaptation function

$H_3(t_r,f_a;R_G)=\exp \left[j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(f_a;R_G)\right]\exp (-j2\mathrm{\π}{f}_a{\stackrel{\‾}{t_a}}^{*})---\left(24\right)$

Data to carrying out inverse Fourier transform, are transformed into t along distance by the data after pulse pressure of adjusting the distance_r-f_aTerritory, Carry out orientation to pulse pressure then in conjunction with formula (24), finally carry out orientation to inverse Fourier transform, data are transformed into t_r-t_aTerritory, finally realizes the coupling imaging of ROSAR data.ROSAR Irnaging procedures figure is as shown in Figure 5.

2. experiment simulation

For the effectiveness of verification method, the point target of ROSAR is emulated by the present invention, ROSAR's Systematic parameter is as shown in Table 1.It is mainly used in low latitude scene imaging, in order to simulate in view of ROSAR system The low latitude handling situations of carrier platform, ground level H chooses smaller value.For orientation to index, due to ROSAR Azimuth resolution change with the change of distance, and azimuth resolution is not changed by distance and is affected^[8], because of This choose azimuth resolution as orientation to a major parameter index.

The systematic parameter of table one ROSAR

Parameter	Value	Parameter	Value
				Antenna is to ground level	100m	Rotor length	5m
Angular velocity	6π/s	Pulse recurrence frequency	1400Hz
				Range resolution	0.5m	Azimuth resolution	0.9°

It is at 4000m that simulating scenes center is selected in distance, it is assumed that irradiating scene width is 3000m, according to table The systematic parameter of one calculates that scene is near, the range curvature difference of distal edge is 0.002m, this value much smaller than distance to Resolution, therefore can ignore the space-variant of range curvature.Many in order to verify that the inventive method can correct simultaneously The range curvature of individual point target, chooses 5 point targets, the circle of each point target in different distance and different azimuth Cylindrical coordinates position be (3950m ,-60 °, 0m), (3950m, 60 °, 0m), (4000m, 0 °, 0m), (4050m ,-60 °, 0m), (4050m, 60 °, 0m).Fig. 6 (a) is the image not carrying out range curvature correction, figure 6 (b) is that the decoupling function utilizing the present invention carries out the image that frequency domain curvature correction is later, during curvature correction, and be Distance is upwards multiplied by the decoupling function of formula (20), and the distance choosing scene center is reference distance, above-mentioned school The most only need multiplication operations of frequency domain.It will be appreciated from fig. 6 that the inventive method can correct multiple point rapidly at frequency domain The range curvature of target.

The orientation pulse pressure method that traditional orientation pulse pressure method and the present invention are proposed by Fig. 7, Fig. 8 compares, Point target position in figure is (4000m, 0 °, 0m).Traditional method uses the approximation of Taylor series expansion, Calculate corresponding orientation to frequency modulation rate, then carry out orientation to pulse pressure.As shown in Fig. 7 (a), traditional method can not Realize mating imaging completely to echo data to orientation.Fig. 7 (b) is that employing formula (24) carries out the orientation knot to pulse pressure Really, during Fig. 8 is Fig. 7 (a) and Fig. 7 (b), distance unit is that the orientation of 568 is to profile.As shown in Figure 8, pass The Incomplete matching of system method causes there is higher secondary lobe near main lobe, and the inventive method can greatly reduce this A little amplitude secondary lobes, are effectively realized azimuth match imaging.

The present invention proposes the frequency domain imaging method of ROSAR, is directly carried out echo-signal by principle in phase point Frequency domain transform, preferably remains the spectrum signature of ROSAR echo-signal.According to obtained two-dimensional frequency Expression formula, structure frequency domain decoupling function, distance are to adaptation function and azimuth match function.The present invention also analyzes The range migration impact of ROSAR, when the direction of illumination of antenna beam and the velocity attitude of antenna movement are mutual Time vertical, it is not necessary to consider the impact of range walk, when the distance at image scene center is higher value, general Under understanding and considerate condition, scene center chooses this condition that meets, and therefore can ignore the shadow of range curvature space-variant Ring.Simulation result shows, utilizes decoupling function and adaptation function that the present invention constructs, it is possible to quickly realize distance Curvature correction and effectively overcome orientation to mismatch problems.

The foregoing is only embodiments of the invention, be not limited to the present invention.The present invention can have respectively Plant suitably change and change.All made within the spirit and principles in the present invention any amendment, equivalent, Improve, should be included within the scope of the present invention.

Claims (1)

1. a rotary synthetic aperture radar frequency domain imaging method, it is characterised in that:

The method utilizes principle in phase point directly ROSAR echo-signal to be transformed to two-dimensional frequency, according to obtained Two-dimensional frequency expression formula, structure frequency domain decoupling function, distance to adaptation function and azimuth match function, Carry out range curvature correction and target scene reconstruct again, concretely comprising the following steps of described method:

A) ROSAR initial data is received；

B) orientation FFT is carried out；

C) distance FFT is carried out；

D) H1 range curvature correction and H2 distance coupling are carried out；

E) operation result of step C and step D is carried out the conversion of distance IFFT；

F) H3 orientation coupling is carried out；

G) operation result of step E and F is carried out the conversion of orientation IFFT；

H) ROSAR image is finally given；

Wherein, first described step A particularly as follows: be fixed on radar antenna on helicopter wing, or installs On unmanned plane or the rigid support of lorry, it is assumed that the height of carrier aircraft platform is H, and antenna is with angular velocity ω acts as uniform circular motion along with wing one, periodically launches signal, backscattering echo warp simultaneously Crossing a fixed response time and return to antenna, then store the echo-signal received, antenna is in ground shape An annular is become to irradiate scene, point target P in scene_nGround distance to center of rotation axle is R_G, The a length of L of rotor, orientation is γ to beam angle, and the angle of pitch is β, and pitching is ε, x to beam angle Direction of principal axis faces out with being defined as parallel to, and z-axis direction is defined as the most upwards, and y-axis is perpendicular to X-z-plane, uses cylindrical-coordinate systemWhen antenna phase central rotation to A point, machine The wing is parallel with x-axis, and order anglec of rotation size now is 0 degree, when antenna rotates to other arbitrfary point B, Anglec of rotation size is ω t_a, the position of B point is (L, ω t_a, H), it is assumed that point target P_nPosition be (R_G, ωt_n, 0), then antenna to the distance expression formula of point target is

$R(t_a;R_G)=\sqrt{L^2-2{\mathrm{LR}}_{G}cos\[\mathrm{\ω}(t_a-t_n)\]+R_G^2+H^2}---\left(1\right)$

The signal of radar emission is linear FM signal, then backscattering echo signal through fundamental frequency convert, away from The slow time domain in m-orientation (t when fast_r-t_aTerritory) it is represented by

$\begin{array}{c}{s}_n(t_r,t_a;R_G)=a_r\[t_r-\frac{2R(t_a;R_G)}{c}\]a_a\left(t_a\right)\exp \left\{{\mathrm{j\πK}}_r{\[t_r-\frac{2R(t_a;R_G)}{c}\]}^2\right\}\\ \×\exp \[-j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(t_a;R_G)\]\end{array}---\left(2\right)$

Wherein,Represent distance to window function,Represent that orientation is to window function, K_rFor signal frequency modulation rate, λ For launching the wavelength of signal, c represents the light velocity；

Wherein, described step B to H particularly as follows:

To s_n(t_r, t_a；R_G) carry out orientation to frequency domain (f_aTerritory) conversion, can obtain

$s_n(t_r,f_a;R_G)={\∫}_{t_n-{\mathrm{\α}}_S/2\mathrm{\ω}}^{t_n+{\mathrm{\α}}_S/2\mathrm{\ω}}{s}_n(t_r,t_a;R_G)\exp (-j2{\mathrm{\πf}}_a{t}_a){\mathrm{dt}}_a---\left(3\right)$

Wherein, t_nIt it is point target P_nThe position time point upwards in orientation, α_SIt is that orientation is to corresponding to synthetic aperture S Synthetic aperture angle, α_SExpression formula be

${\mathrm{\α}}_S=\mathrm{\γ}(1-\frac{L}{R_G})---\left(4\right)$

Amplitude in formula (3) is gradual, and phase place is expressed as

By formula (5) to slow time t_aDerivation, and function is 0 after making derivation, can be in the hope of

$sin\[\mathrm{\ω}(t_a^{*}-t_n)\]=-\frac{\mathrm{\λ}}{2{\mathrm{LR}}_G\mathrm{\ω}}{f}_a\sqrt{L^2+R_G^2+H^2-\[\frac{{\mathrm{\λ}}^2}{2{\mathrm{\ω}}^2}{f}_a^2+\sqrt{\frac{{\mathrm{\λ}}^4}{4{\mathrm{\ω}}^4}{f}_a^4-\frac{{\mathrm{\λ}}^2}{{\mathrm{\ω}}^2}(L^2+R_G^2+H^2)f_a^2+4L^2{R}_G^2}\]}---\left(6\right)$

Wherein,It is in phase point, as ω (t_a-t_nDuring)=pi/2, oblique distance isAnd makeThen (6) can be simplified, and then obtain

$\begin{array}{c}{t}_a^{*}=t_n-\frac{1}{\mathrm{\ω}}\arcsin \[\frac{1}{2{\mathrm{LR}}_G}{\stackrel{\‾}{f}}_a\sqrt{R_{\max }^2-(\frac{1}{2}{\stackrel{\‾}{f}}_a^2+\sqrt{\frac{1}{4}{\stackrel{\‾}{f}}_a^4-R_{\max }^2{\stackrel{\‾}{f}}_a^2+4L^2{R}_G^2})}\]\\ =t_n-{\stackrel{\‾}{t}}_a^{*}\end{array}---\left(7\right)$

According to principle in phase point, formula (7) is substituted into formula (3), can obtain

$\begin{array}{c}{s}_n(t_r,f_a;R_G)=a_r\[t_r-\frac{2R(f_a;R_G)}{c}\]a_a\left(t_a^{*}\right)\exp \left\{{\mathrm{j\πK}}_e(f_a;R_G){\[t_r-\frac{2R(f_a;R_G)}{c}\]}^2\right\}\\ \×\exp \[-j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(f_a;R_G)\]\exp \left(j2{\mathrm{\πf}}_a{\stackrel{\‾}{t}}_a^{*}\right)\exp (-j2{\mathrm{\πf}}_a{t}_n)\end{array}---\left(8\right)$

Wherein, R (f_a；R_G) it is that oblique distance R is at f_aExpression formula on territory, works as t_aTerritory transforms to f_aDuring territory, launch signal Frequency modulation rate K_rAlso can change, make K_rAt f_aK it is expressed as on territory_e(f_a；R_G), be given individually below R(f_a；R_G) and K_e(f_a；R_G) expression,

Formula (7) is substituted into formula (1), tries to achieve R (f_a；R_G) expression formula be

$R(f_a;R_G)=\sqrt{R_{max}^2-(\frac{1}{2}{\stackrel{\‾}{f}}_a^2+\sqrt{\frac{1}{4}{\stackrel{\‾}{f}}_a^4-R_{max}^2{\stackrel{\‾}{f}}_a^2+4L^2{R}_G^2})}---\left(9\right)$

Work as f_aWhen=0, R (f can be obtained_a；R_G) constant term beThis constant term It it is point target P_nFormula (9) therefore can be rewritten as by the beeline in oblique distance face

R(f_a；R_G)=R_C+ΔR(f_a；R_G) (10)

Wherein, Δ R (f_a；R_G) it is R (f_a；R_G) once and high-order term,

K can be obtained by analysis below_e(f_a；R_G), at t_r-t_aIn territory, the echo data of a single point target is along fast Time t_rVertical distribution, on certain string storage position, works as t_aTerritory transforms to f_aDuring territory, the storage position of echo data Put and become skew lines, be f for tranmitting frequency_cSignal, antenna rotates to a certain angle, θ_t=ω t_a, now Doppler values be

$F_a=-\frac{2{\mathrm{LR}}_G\mathrm{\ω}sin\left({\mathrm{\θ}}_t\right)f_c}{cR\left({\mathrm{\θ}}_t\right)}---\left(11\right)$

Wherein, R (θ_t) it is the R (t in formula (1)_a；R_G), it is linear FM signal owing to launching signal, when launching frequency Rate becomes f=f_c+ Δ f (Δ f=K_rΔt_r), anglec of rotation θ_tCorresponding doppler values is

$F_a^{\′}=-\frac{2{\mathrm{LR}}_G\mathrm{\ω}sin\left({\mathrm{\θ}}_t\right)(f_c+\mathrm{\Δ}f)}{cR\left({\mathrm{\θ}}_t\right)}---\left(12\right)$

Now, due to the change of tranmitting frequency, storage position is transferred to a little 2 by putting 1, i.e. for a single point target Saying, the storage position of echo data is at t_r-f_aIt territory is skew lines；

It is f for tranmitting frequency_c+ Δ f signal, doppler values F_aCorresponding storage data are another echoes Received data, the echo data that aerial position B ' is received, now the anglec of rotation is θ_t-Δθ_t, phase The doppler values expression formula answered is

$F_a=-\frac{2{\mathrm{LR}}_G\mathrm{\ω}sin({\mathrm{\θ}}_t-{\mathrm{\Δ\θ}}_t)(f_c+\mathrm{\Δ}f)}{cR({\mathrm{\θ}}_t-{\mathrm{\Δ\θ}}_t)}---\left(13\right)$

Understand, due to t according to above analysis_aTerritory transforms to f_aTerritory, the transformation required for same frequency difference DELTA f Time, on same slow time point, tranmitting frequency was by f by different_cTransform to f_c+ Δ f required time is Δt_r(Δt_r=Δ f/K_r), but for same doppler values F_a, tranmitting frequency is by f_cTransform to f_c+ Δ f institute Take time Δ t '_r(Δt′_r=Δ f/K_e(f_a；R_G)) it is

${\mathrm{\Δt}}_r^{\′}={\mathrm{\Δt}}_r-\frac{2\mathrm{\Δ}R\left({\mathrm{\θ}}_t\right)}{c}---\left(14\right)$

Wherein, Δ R (θ_t)=R (θ_t)-R(θ_t-Δθ_t), in order to eliminate intermediate variable Δ f, need Δ R (θ_t) be expressed as The expression formula of Δ f, can obtain according to formula (1)

$\mathrm{\Δ}R\left({\mathrm{\θ}}_t\right)=R\left({\mathrm{\θ}}_t\right)-R({\mathrm{\θ}}_t-{\mathrm{\Δ\θ}}_t)=\frac{{\mathrm{LR}}_G}{R\left({\mathrm{\θ}}_t\right)}sin\left({\mathrm{\θ}}_t\right)\sin \left({\mathrm{\Δ\θ}}_t\right)---\left(15\right)$

Can obtain according to formula (11) and formula (13) again

$sin\left({\mathrm{\Δ\θ}}_t\right)=tan\left({\mathrm{\θ}}_t\right)\[\frac{\mathrm{\Δ}R\left({\mathrm{\θ}}_t\right)}{R\left({\mathrm{\θ}}_t\right)}+\frac{\mathrm{\Δ}f}{f_c}\]---\left(16\right)$

Formula (16) is substituted into formula (15), it is thus achieved that Δ R (θ_t) about the expression formula of Δ f be

$\mathrm{\Δ}R\left({\mathrm{\θ}}_t\right)=\frac{{\mathrm{LR}}_{G}sin\left({\mathrm{\θ}}_t\right)tan\left({\mathrm{\θ}}_t\right)}{R^2\left({\mathrm{\θ}}_t\right)-{\mathrm{LR}}_{G}sin\left({\mathrm{\θ}}_t\right)tan\left({\mathrm{\θ}}_t\right)}\frac{\mathrm{\Δ}f}{f_c}R\left({\mathrm{\θ}}_t\right)---\left(17\right)$

Variable in formula (17) is θ_t(θ_t=ω t_a), t in formula to be utilized (7)_aWith f_aRelation, by formula (17) Transform to f_aTerritory, corresponding θ_tIt is transformed into θ_f, finally utilize formula (14) and formula (17), after eliminating intermediate variable Δ f Can obtain

Due to Δ t_r=Δ f/K_r, Δ t '_r=Δ f/K_e(f_a；R_G), can obtain according to formula (16) and formula (17)

$\frac{1}{K_e(f_a;R_G)}=\frac{1}{K_r}-\frac{2{\mathrm{LR}}_{G}sin\left({\mathrm{\θ}}_f\right)tan\left({\mathrm{\θ}}_f\right)}{R^2\left({\mathrm{\θ}}_f\right)-{\mathrm{LR}}_G\sin \left({\mathrm{\θ}}_f\right)tan\left({\mathrm{\θ}}_f\right)}\frac{\mathrm{\λ}}{c^2}R\left({\mathrm{\θ}}_f\right)---\left(18\right)$

So, oblique distance and frequency modulation rate are just obtained at f_aExpression formula on territory, by the R (f of formula (10)_a；R_G) and formula (18) K_e(f_a；R_G) substitute into formula (8), i.e. can get echo-signal at t_r-f_aExpression formula on territory s_n(t_r, f_a；R_G)；

After orientation is to frequency domain transform, recycle principle in phase point, to s_n(t_r, f_a；R_G) carry out distance to Frequency domain (f_rTerritory) conversion, can be in the hope of

$\begin{array}{c}{s}_n(f_r,f_a;R_G)=a_r\[\frac{f_r}{K_e(f_a;R_G)}\]a_a\left(t_a^{*}\right)\exp \[-j\mathrm{\π}\frac{f_r^2}{K_e(f_a;R_G)}\]\\ \×\exp \[-j\frac{4\mathrm{\π}}{c}\mathrm{\Δ}R(f_a;R_G)f_r\]\exp (-j\frac{4\mathrm{\π}}{c}{R}_C{f}_r)\\ \×\exp \[-j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(f_a;R_G)\]\exp \left(j2{\mathrm{\πf}}_a{\stackrel{\‾}{t}}_a^{*}\right)\exp (-j2{\mathrm{\πf}}_a{t}_n)\end{array}---\left(19\right)$

Above formula is the two-dimensional frequency expression formula of ROSAR echo-signal；

Second exponential term in formula (19) is the azimuthrange coupling item caused by range curvature, therefore correct away from Decoupling function from bending is

$H_1(f_r,f_a;R_G)=\exp \[j\frac{4\mathrm{\π}}{c}\mathrm{\Δ}R(f_a;R_G)f_r\]---\left(20\right)$

Wherein, Δ R (f_a；R_G) degree of crook with R_GChange and change, i.e. oblique distance R is at f_aSpace-variant is there is in territory Property；

The synthetic aperture angle provided according to formula (4), can be in the hope of a certain distance R_GCorresponding is the most general Le value is

$f_{amax}=-\frac{2{\mathrm{LR}}_G\mathrm{\ω}sin\left(\frac{{\mathrm{\α}}_S}{2}\right)}{\mathrm{\λ}\sqrt{L^2-2{\mathrm{LR}}_{G}cos\left(\frac{{\mathrm{\α}}_S}{2}\right)+R_G^2+H^2}}---\left(21\right)$

Can be in the hope of oblique distance R at f according to formula (10) and formula (21)_aThe ultimate value of maximum deflection value in territory

$\underset{R_G\→\mathrm{\∞}}{\lim }\mathrm{\Δ}R(f_{amax};R_G)=L\[1-cos\left(\frac{\mathrm{\γ}}{2}\right)\]---\left(22\right)$

The distance choosing scene center is reference distance, then range curvatures all in scene is unified school Just；

After range curvature correction, in order to carry out distance to pulse pressure, distance adaptation function need to be constructed as follows

$H_2(f_r,f_a;R_G)=\exp \[j\mathrm{\π}\frac{f_r^2}{K_e(f_a;R_G)}\]---\left(23\right)$

Original fundamental frequency echo data is carried out bidimensional frequency domain transform, data are transformed into f_r-f_aTerritory, then utilizes Formula (20) and formula (23) realize range curvature correction and distance to pulse pressure, in order to complete orientation to pulse pressure, structure side Position is as follows to adaptation function

$H_3(t_r,f_a;R_G)=\exp \[j\frac{4\mathrm{\π}}{\mathrm{\λ}}R(f_a;R_G)\]\exp (-j2{\mathrm{\πf}}_a{\stackrel{\‾}{t}}_a^{*})---\left(24\right)$

Data to carrying out inverse Fourier transform, are transformed into t along distance by the data after pulse pressure of adjusting the distance_r-f_aTerritory, Carry out orientation to pulse pressure then in conjunction with formula (24), finally carry out orientation to inverse Fourier transform, data are transformed into t_r-t_aTerritory, finally realizes the coupling imaging of ROSAR data.